# Question 31ddd

Jul 12, 2017

68.9461%

#### Explanation:

Equation for the reaction:
$F {e}_{3} {O}_{4} + {H}_{2} \to {H}_{2} O + F e$

Balanced equation:
$F {e}_{3} {O}_{4} + 4 {H}_{2} \to 4 {H}_{2} O + 3 F e$

Theoretically,
1 mol of $F {e}_{3} {O}_{4}$ reacts with 4 mol of ${H}_{2}$ to yield 4 mol of ${H}_{2} O$

Calculating the mole of $F {e}_{3} {O}_{4}$ and ${H}_{2}$

mole$= \left(\text{mass"("g"))/("molar mass"("g"/"mol}\right)$

Molar mass of a compound is the sum of each atom molar mass in a molecule of the compund.
Obtain the molar mass of elements from a periodic table

molar mass of $F {e}_{3} {O}_{4} = \left[55.845 \cdot 3 + 15.999 \cdot 4\right] \left(\frac{g}{m o l}\right)$
molar mass of $F {e}_{3} {O}_{4} = 231.531 \text{g/"mol}$

molar mass of ${H}_{2} = 1.008 \cdot 2 \text{g"/"mol}$
molar mass of ${H}_{2} = 2.016 \text{g"/"mol}$

mol of $F {e}_{3} {O}_{4} = \frac{963.1}{231.531} = 4.1597 \text{mol}$
mol of ${H}_{2} = \frac{8.57}{2.016} = 4.251 \text{mol}$

which is the limiting reactant?

since 4 mol of ${H}_{2}$ (biggest mol number) is needed for the reaction
divide the mol of ${H}_{2}$ by 4 to find the respective mol of $F {e}_{3} {O}_{4}$ to react completely with all of the ${H}_{2}$ available
("mol of "H_2)/4=4.251/4"mol"=1.06275"mol"

${H}_{2}$ is the limiting reactant, $F {e}_{3} {O}_{4}$ is said to be in excess

find the mol of water produced
4 mol of ${H}_{2}$ : 4 mol of ${H}_{2} O$
$\therefore$4.251mol of ${H}_{2}$ : 4.251 mol of ${H}_{2} O$

PERCENTAGE YIELD
PERCENTAGE YIELD=("experimental value")/("actual value")*100%
the value could be mass/mol/volume...

In this case I will use mol as the value.

We need experimental mol of ${H}_{2} O$

molar mass of ${H}_{2} O = \left[1.008 \cdot 2 + 15.999\right] \text{g"/"mol}$
molar mass of ${H}_{2} O = 18.015 \text{g"/"mol}$

mol of ${H}_{2} O = \frac{52.8}{18.015} = 2.9309 \text{mol}$

PERCENTAGE YIELD =2.9309/(4.251)*100%=68.9461%#